1. ## Confused About Drawing With X Intercepts

For each of the following:

y= A ^ (x-2) - 3

if A> 1

domain:
range:
asymptote:

Then Sketch.

i got domain = all real
range= y> -3 and asymptote -3.
Im not sure if this is right though..
and i need to draw a sketch of this..
no idea how to. could some1 please explain how to do this.

thanks.

2. Originally Posted by clp_53
For each of the following:

y= A ^ (x-2) - 3

if A> 1

...
i got domain = all real ....... OK
range= y> -3 ....... OK
and asymptote: y = -3. if x approaches negative infinity.
Im not sure if this is right though..
and i need to draw a sketch of this..
no idea how to. could some1 please explain how to do this.

thanks.
All graphs of the family of functions pass through P(2, -2). (Why?)

I've attached a sketch of the graphs of:

$y = \left(\frac32\right)^{x-2} - 3$ and

$y = 2^{x-2} - 3$ and

$y = 3^{x-2} - 3$

3. Hello, clp_53!

$y\:=\:A ^ {x-2} - 3\quad A> 1$

Find: the domain, range, and asymptote(s) ... then sketch.

i got: . $\begin{Bmatrix}\text{Domain:} & \text{all real x} \\\text{Range:} & y \,>\, -3 \\\text{Asymptote:} & y = -3 \end{Bmatrix}$ . . . . Right!
We're expected to know the graph of $y \:=\:A^x,\;\;A > 1$
Code:
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It has y-intercept (0,1).
To the left, it approaches the x-axis.
To the right, it soars upward.

$y \:=\:A^{x{\color{red}-2}}$ .is the same graph moved 2 units to the right.

Then: . $y \:=\:A^{x-2} \,{\color{red}-\, 3}$ .moves it 3 units down.

Got it?